SELECTED DISCUSSIONS TO exercises/problems
BELOW.
YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE
DESIGNED TO HELP YOU DO RELATED EXERCISES. WE WILL REVIEW THEM IN-CLASS.
ALSO READ ALL EXAMPLES; THE EXERCISES ARE SIMILAR.

CH. 24

CH. 24 MULTIPLE CHOICE TBA -

TURN IN: CH. 24---5, 6, 10, 12, 13, 14, 29,
30, 33, 34, 50, 52, 53

* DISCUSSIONS PROVIDED.

5. Use 1/f = 1/di + 1/do and f = R/2
> 0. Find di and m = -di/do, from which you may get final
height hf = m*hi; note hf is negative if initial height hi is positive
if the image is inverted and real. That's because di > 0 and m = -di/do
< 0. That happens when do > f. When do < f, the image is
virtual and upright and di <0, making m = -di/do > 0. See # 12
below.

6. Let f = -|R|/2. Use 1/f = 1/di +
1/do. Find di and m = -di/do, from which you may get final height
hf = m*hi; note hf is positive if initial height hi is positive since
the image is upright and virtual . Note that di < 0 in all cases.
Thus m > 0.

12. (a) Use SYMBOLS: Use 1/f = 1/di
+ 1/do and f > 0. Find di and m = -di/do if do > f. Show di
> 0 from which you get final height hf = m*hi; note hf is
negative if initial height hi is positive since the image is inverted
and real. That's because m = -di/do < 0.
(b) Use SYMBOLS: Use 1/f = 1/di + 1/do and f > 0. Find di
and m = -di/do if do < f. Show di < 0 from which you
get final height hf = m*hi; note hf is positive if initial height hi is
positive since the image is upright and virtual. That's because m
= -di/do is now positive!

13. Use 1/f = 1/di + 1/do and f =
R/2 > 0. Find di and m = -di/do, from which you may get final
height hf = m*hi; note hf is negative if initial height hi is positive
if the image is inverted and real . That happens when do > f.
When do < f, the image is virtual and upright and di <0, making m
= -di/do > 0. See # 12 above.
(a) Find m.
(b) Find di.
(c) See distributed class notes where we drew strategic rays to map out,
through ray tracing, the location of the image given an
object, its distance from mirror do and height ho. In
this case you are given do but no number for ho. Still you can draw an
arrow at the location of the object, distance do from mirror surface at
axis. The arrow tail is on the axis line of mirror and the tip
is vertically above that point some finite distance defining arrow's
height ho.

14. Use 1/f = 1/di + 1/do and f =
R/2 > 0. Find di and m = -di/do, from which you may get final
height hf = m*hi; note hf is negative if initial height hi is positive
if the image is inverted and real . That happens when do > f.
When do < f, the image is virtual and upright and di <0, making m
= -di/do > 0. See # 12 above.
(a) See distributed class notes where we drew strategic rays to
map out, through ray tracing, the location of the image given an
object, its distance from mirror do and height ho. In
this case you are given do and number for ho. Still you can
draw an arrow at the location of the object, distance do from mirror
surface at axis. The arrow tail is on the axis line of mirror and
the tip is vertically above that point some finite distance
defining arrow's height ho..
(b) Find di, m and hf given ho > 0. Is di > 0 or di < 0? Is the
image real (inverted) or real (upright); in other words is m < 0 or m
>0, where m = - di/do?

29. Use 1/f = 1/di + 1/do and
f > 0. Find di and m = -di/do, from which you may get
final height hf = m*hi; note hf is negative if initial height hi is
positive if the image is inverted and real . That happens when do
> f. When do < f, the image is virtual and upright and di <0,
making m = -di/do > 0.

The object is to the left of lens since the virtual image
is stated to be upright like the object;see
fig. 24. 37 (e). You have two unknowns, di and do,
and two equations:
Find (i)1/f = 1/di + 1/do , where f > 0
and (ii) hf = -(di/do)*hi = m*hi, where m = -di/do.
(ii) may be re-written and di = -(1.30 cm /0.400 cm)*do or
(ii) di = -3.25do .

30. Use 1/f = 1/di + 1/do and
f > 0. Find di and m = -di/do, from which you may get
final height hf = m*hi; note hf is negative if initial height hi is
positive if the image is inverted and real . That happens when do
> f. When do < f, the image is virtual and upright and di <0,
making m = -di/do > 0.

The object is to the right of lens since the real image is
stated to be inverted (unlikethe object) and on the left
of the lens; see figs. 24. 37 (a), (b), (c); you might
have to rotate the figures to fit this problem such that object is
to right of lens.